const-typed class does not have const members

[Ali Çehreli]

Sean Eskapp wrote:
 > The following code yields results as commented.
 >
 > import std.stdio;
 >
 > class A
 > {
 > 	int b;
 > }
 >
 > void main()
 > {
 > 	const A a = new A;
 > 	writeln(typeof(a).stringof); // const(A)
 > 	writeln(typeof(a.b).stringof); // const(int)
 >
 > 	writeln((const A).stringof); // const(A)
 > 	writeln(typeof((const A).b).stringof); // int
 > }
 >
 > Given this information, how do I work around this in templates? I'd 
like a
 > template which can type its params based on members of T, for instance:
 >
 > void foo(T)(T x, typeof(T.member) y)
 >
 > but this will yield non-const second parameters even if T is const. 
Help anybody?

Ironically, parameter list is not a useful place to constraint the 
parameter types. At the least, you probably want some conversions. If 
so, would this work:

import std.traits;

void foo(T, M)(T x, M m)
     if (isImplicitlyConvertible!(M, typeof(T.member)))
{}

class C
{
     int member;
}

void main()
{
     auto c = new C;
     foo(c, 42);
}

If you want to ensure that the member is assignable, you can try this:

import std.traits;

void foo(T, M)(T x, M m)
     if (__traits(compiles, { x.member = m; }))
{}

class C
{
     int member;
}

void main()
{
     const auto c = new C;  // <-- compilation error
     foo(c, 42);
}

But the error message is only somewhat helpful:

deneme.d(26369): Error: template deneme.foo(T,M) if 
(__traits(compiles,delegate ()
{
x.member = m;
}
)) does not match any function template declaration
deneme.d(26369): Error: template deneme.foo(T,M) if 
(__traits(compiles,delegate ()
{
x.member = m;
}
)) cannot deduce template function from argument types !()(const(C),int)

Ali